1. Write the reaction

$3\,HC\equiv CH_{(g)} \rightleftharpoons C_6H_6{}_{(l)}$

We use

$\Delta G^\circ_{\text{rxn}} = \sum \nu \Delta_f G^\circ(\text{products}) - \sum \nu \Delta_f G^\circ(\text{reactants})$

2. Calculate $\Delta G^\circ_{\text{rxn}}$

Given:

$\Delta_f G^\circ(HC\equiv CH) = -2.04\times 10^5\; \text{J mol}^{-1}$ $\Delta_f G^\circ(C_6H_6) = -1.24\times 10^5\; \text{J mol}^{-1}$

So,

$\Delta G^\circ_{\text{rxn}} = \left(-1.24\times 10^5\right) - 3\left(-2.04\times 10^5\right)$

$= -1.24\times 10^5 + 6.12\times 10^5$

$= 4.88\times 10^5\; \text{J mol}^{-1}$

3. Use the relation between $\Delta G^\circ$ and $K$

$\Delta G^\circ = -RT\ln K$

Hence,

$\ln K = -\frac{\Delta G^\circ}{RT}$

At $25^\circ C = 298\,K$,

$\ln K = -\frac{4.88\times 10^5}{8.314\times 298}$

First compute denominator:

$8.314\times 298 = 2477.572$

Thus,

$\ln K \approx -\frac{4.88\times 10^5}{2477.572} \approx -196.97$

4. Convert to common logarithm

$\log K = \frac{\ln K}{2.303}$

So,

$\log K \approx \frac{-196.97}{2.303} \approx -85.53$

Therefore, the magnitude of $\log K$ is

$|\log K| \approx 85.53$

5. Match with the form given

The question says magnitude of $\log K$ is $x\times 10^{-1}$.

So,

$x\times 10^{-1} = 85.53$

$x = 855.3$

Since integer answer is required,

$x = 855$

6. Comparison with stored answer

Stored correct answer = $855$

This matches our derived answer.