1. Use the relation between standard Gibbs energy and equilibrium constant

For the reaction $2A(g) \rightleftharpoons A_2(g)$ we use: $\Delta G^\circ = -RT \ln K$

Since the question asks for $K_C$, and for this reaction in gas phase we first find $K_P$ from $\Delta G^\circ$.

2. Calculate $\ln K_P$

Given: $\Delta G^\circ = +25.2\,\text{kJ mol}^{-1} = 25200\,\text{J mol}^{-1}$ $R = 8.3\,\text{J mol}^{-1}\text{K}^{-1}, \quad T=400\,\text{K}$

So, $\ln K_P = -\frac{\Delta G^\circ}{RT} = -\frac{25200}{8.3\times 400}$

$8.3\times 400 = 3320$

$\ln K_P = -\frac{25200}{3320} \approx -7.59$

Now convert to base 10: $\log K_P = \frac{\ln K_P}{2.3} = \frac{-7.59}{2.3} \approx -3.30$

Thus, $K_P = 10^{-3.30}$

Using $10^{-3.30} = 10^{-3}\times 10^{-0.30}$ and $\text{antilog}(-0.3)=0.501$

we get: $K_P = 10^{-3}\times 0.501 = 5.01\times 10^{-4}$

3. Relate $K_P$ and $K_C$

For $2A(g) \rightleftharpoons A_2(g)$

$\Delta n = 1-2 = -1$

Relation: $K_P = K_C (RT)^{\Delta n} = K_C (RT)^{-1}$

Hence, $K_C = K_P\times RT$

Given that $1\,\text{atm}=1\,\text{bar}$, use $RT = 0.083\times 400 = 33.2$

Therefore, $K_C = 5.01\times 10^{-4}\times 33.2$

$K_C \approx 1.66\times 10^{-2}$

4. Match with required form

The question asks: $K_C = \underline{\hspace{1cm}}\times 10^{-2}$

So the coefficient is $1.66 \approx 2$

5. Final Answer

$K_C \approx 2\times 10^{-2}$

So the required integer is: $\boxed{2}$