Calculate capacitor to hold up current limited supply
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.