Well, first of all there would be a shockwave. In order to calculate the blast effect you would need to know the yield of the bomb which would then convert from the usual measure of kilotons to the more familiar to the typical physics problem measure of Joules. I think that there is some standard that 1kiltoton = 4.184*10^12 Joules; but I am not sure of this without looking up the exact number. Assuming that you have a given quantity of "heat", imagine that the bomb is vaporized - which it woud be, so use some of the familiar equations like PV=nRT to calcuate some meaningful number for the temperature and pressure of the resulting ball of plasma that will result. Then calculate the free exapansion of this ball of plasma into a sphere of 400 meters radius. That will give you a number for the temperature and pressure of the expanding plasma.
Now for the hint: When a volume of gas expands against a piston; or otherwise has something to do "work" on, you have to do some more exotic calculations; like "the work done by the adiabatic expansion of a gas", and for an atmospheric blast this implies that the expanding fireball does "work" on the atmosphere; and this is what causes much of the "heat and blast" effect, just like a diesel engine, i.e. the expanding fireball heats the atmosphere not simply because of the thermal effects of the explosion, but rather you have a heating effect that propagates at the speed of sound because of the compression of the air. This is in addition to the effect of radiation ionizing the atmosphere - dont confuse the two!
Now in a vacuum of course you aren't ionizing the atmosphere; you simply have a certain amount of energy given of as radiation, so the so called "radiation transport" piece goes away. That is to say that with an air burst of a nuclear device - radiation is absorbed by the atmosphere as gamma rays; then the atmosphere emits x-rays, which in turn are absorbed and re-emitted as ultraviolet and so on down the chain. This is the so-called "radiation transport" and for an atmospheric blast it makes the computations very difficult to actually perform, that is because you would have to take into account the absorption and emission spectra of the various plasmas that can be made from the atmosphere.
Now the good news finally, if there is any is that you dont have to deal with that in a vacuum, you only have to cacluate the intensity of the energy that reaches your space ship - it will all pretty much be in the form of radiation and shock wave of the expanding plasma cloud.
So lets see PV=nRT implies P=nRT/V. With V=(4*pi*(400m)^3)/3. Now to come up with reasonable values for the mass of the device, including all shielding - that is to say that we are assuming that the bomb completely vaporizes ...
Right if if ALL vaporizes ... but if it doesnt, I mean if so much as some grain of sand size particles survive the blast and are driven at high speed in the direction of the space ship - well it should be very obvious that that would be very bad for the space ship. Noting that the "peace" dome in Hiroshima was not "completely" vaporized by the blast, seems to strongly suggest that in space a nuclear bomb might still be usable as the core of a very nasty "grendade" -- lots of potential for microscopic "shrapnel" to inflict great harm to nearby space craft, or even satellites thousands of miles away.
My "guess" is leaning toward various other factors such as super heated micrometeorites moving many miles per second will pulverize any heat shielding, or "tiles" or anything else they run into. The velocity of these micrometeorites could very well be high enough to send some into higher orbits and damage or destroy orbiting satellites thousands of miles away.
I dont have the total energy from the "shock wave" but as described, it is very real - no matter what you hear about there being "no sound in space" and no matter how empty you think the space was before the blast .... so lets make a guess on the shock wave energy:
For a 10kT device, lets assume a total energy of 4*10^13 Joules; and divide that by 4*pi*(400meters)^2 - I am getting about 20 megajoules per square meter in "Windows Calculator" - the back of your envelope may vary. Now once again, based on the laws of physics -- all of this energy will be coming at your spaceship, it will be in the form of microscopic shrapnel that survives the blast, but mostly in the form of radiation and the plasma fireball that will carry much of the energy in the combined "heat and shock" of the blast. For the reasons explained, i.e. the lack of "radiation transport" over an extended distance - most of what you think of as a traditional blast, mushroom cloud, etc, wont be present.