Explanation :
$ \begin{array}{l}{\text{KVA}}_1=\sqrt{800^2+600^2}\\\;\;\;\;\;\;\;\;\;=1000\;KVA\end{array} $
Without excessive heat dissipation means current should be constant (i.e.) KVA erating must be constant.
In second case Active power,
$ \mathrm P=800+100=900\;\mathrm{KW} $
Reactive power in second case,
$ \begin{array}{l}{\mathrm Q}_2=\sqrt{1000^2-900^2}\\\;\;\;\;=435.889\;\mathrm{KVAR}\end{array} $
Reactive power supplied by the three phase bank
$ \begin{array}{l}=600-435.889\\\;\;\;\;\;\;\;\;\;\;\;\;\;=164.11\;\mathrm{KVAR}\\{\mathrm Q}_\mathrm{bank}/\mathrm{ph}=\frac{164.11}3=54.7\;\mathrm{KVAR}\\\;\;\;\;\;\;\mathrm V/\mathrm{ph}=\frac{3.3}{\sqrt3}=1.9052\;\mathrm{KV}\\\;\;\;\;\mathrm{Qc}/\mathrm{ph}=\frac{\left(\mathrm{VI}/\mathrm{ph}^2\right)}{{\mathrm X}_\mathrm C}\\\;\;\;\;\;\;\;\;\;\;{\mathrm X}_\mathrm C=\frac{\left(1.9052\times10^3\right)^2}{54.7\times10^3}=66.36\;\mathrm W\\\;\;\;\;\;\;\;\;\;\;\;\;\mathrm C=\frac1{2{\mathrm{πfX}}_\mathrm C}=\frac1{2\mathrm\pi\times50\times66.36}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=47.96\end{array} $