To determine the change in standard Gibbs free energy ($\Delta G^{\circ}$) for the reaction at a given temperature when the equilibrium constant ($K$) is known, we can use the following relationship:

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

Here, $R$ is the universal gas constant ($8.314 \ J K^{-1} mol^{-1}$), $T$ is the temperature in Kelvin ($300 \ K$), and $K$ is the equilibrium constant.

Puting the values we get,

$$ \Delta G^{\circ} = -8.314 \times 300 \times \ln(10) $$

We can convert the natural logarithm of 10 to base-10 logarithm, using the change of base formula, $\ln(10) = 2.3026$. So,

$$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 $$

Calculating this yields:

$$ \Delta G^{\circ} = -8.314 \times 300 \times 2.3026 = -5743.914 \ J mol^{-1} $$

To convert this to kilojoules per mole, we divide by 1000:

$$ \Delta G^{\circ} = -5.743914 \times 10^{3} \ J mol^{-1} \times \frac{1 \ kJ}{10^{3} J} = -5.743914 \ kJ mol^{-1} $$

Now, when expressing this in terms of $\times 10^{-1} \ kJ mol^{-1}$, we get:

$$ \Delta G^{\circ} = -57.43914 \times 10^{-1} \ kJ mol^{-1} $$

Therefore, $\Delta G^{\circ}$ for the reaction is approximately $-57.43914 \times 10^{-1} \ kJ mol^{-1}$ or $-57.4 \times 10^{-1} \ kJ mol^{-1}$ when rounded to three significant figures.