For reaction: ${\mathrm{SO}}_2( \mathrm{g})+\frac{1}{2}{\mathrm{O}}_2( \mathrm{g})\rightleftharpoons {\mathrm{SO}}_3( \mathrm{g})$ ${\mathrm{K}}_{\mathrm{P}}=2\times {10}^{12}$ at $27°\mathrm{C}$ and $1 \mathrm{atm}$ pressure. The ${\mathrm{K}}_{\mathrm{c}}$ for the same reaction is $\times {10}^{13}.($ Nearest integer)
(Given $\mathrm{R}=0.082 \mathrm{L} \mathrm{atm} {\mathrm{K}}^{-1} {\mathrm{mol}}^{-1}$ )

For reaction: ${\mathrm{SO}}_{2\left(\mathrm{g}\right)} + \frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)} \rightleftharpoons {\mathrm{SO}}_{3\left(\mathrm{g}\right)}$

${\mathrm{K}}_{\mathrm{p}} = 2\times {10}^{12}$ at $300 \mathrm{K}$ and $1 \mathrm{atm}$ pressure.

The relation between equilibrium constant in terms of pressures ${\mathrm{K}}_{\mathrm{p}}$ and equilibrium constant in terms of concentration ${\mathrm{K}}_{\mathrm{c}}$ is

${\mathrm{K}}_{\mathrm{p}} = {\mathrm{K}}_{\mathrm{c}} \times (\mathrm{RT}{)}^{∆ng}$

$\Rightarrow 2\times {10}^{12} = {\mathrm{K}}_{\mathrm{c}} \times (0.082 \times 300{)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow {\mathrm{K}}_{\mathrm{C}}= 9.92 \times {10}^{12}$

$\Rightarrow {\mathrm{K}}_{\mathrm{C}} = 0.992 \times {10}^{13}$

So, answer is 1.