Electronic – Hold Up for Power loss of 10ms

For a capacitor, the charge \$Q\$ is proportional to the voltage \$V\$. \$Q = C\times V\$. Rearrange, \$C=\frac{Q}{V}\$.

Charge is current multiplied by time. \$Q = 1.8\,\mathrm{A} \times 15\,\mathrm{ms} = 0.027\,\mathrm{C}\$.

Voltage is \$150\,\mathrm{V}\$, from your question description.

So \$C = \frac{0.027}{150} = 180\,\mu\mathrm{F}\$.

Of course, this assumes the power supply contributes zero current during the pulse, and that the current of the pulse is a constant \$1.8\,\mathrm{A}\$. That doesn't match up with your circuit, which shows a fixed \$2.8\,\mathrm{kV}\$ source and a \$2\,\mathrm{k}\Omega\$ resistor, which will draw \$1.4\,\mathrm{A}\$, and the capacitor will do nothing. As Arsenal rightly points out in his comment, you should be concerned as to how your supply will actually react with that \$2\,\mathrm{k}\Omega\$ there.

You are more or less close to the answer, but you should rather use the capacitor energy equation: \$E = \frac{1}{2} C * V^2\$

The Energy that you need to store can be calculated as

\$E_{needed} = Power * time = ( 12 V * 10 A ) * 10 ms\$

Then you have two instants of time, before and after the switching:

\$E_{before} = \frac{1}{2} C * V_{before}^2\$

\$E_{after} = \frac{1}{2} C * V_{after}^2\$

then \$E_{needed} = E_{before} - E_{after}\$

Also you want \$V_{after}\$ and \$V_{before}\$ close to each other, perhaps 1 V is ok (you must check the load specifications), so there is enough voltage to feed the load, \$V_{before}=12 V\$, \$V_{after}=11 V\$.

The voltage will always go down a "little", that is the rule when discharging capacitors, how "little" depends on how big the capacitor.