Step 1:

Let $E_1,E_2$ and $E_3$ be the event that boxes I,II and III are chosen respectively.

The $P(E_1)=P(E_2)=P(E_3)=\large\frac{1}{3}$

Let the event A be the event that the coin drawn in of gold.

Then $P(A/E_1)$=P(a gold coin from bag I)=$\large\frac{2}{2}$=1

$P(A/E_2)$=P(a gold coin from bag II)=0

$P(A/E_3)$=P(a gold coin from bag III)=$\large\frac{1}{2}$

Step 2:

The probability that the other coin in the box is of gold=the probability that gold coin is drawn from the box I

$\Rightarrow P(E_1/A)$

By Baye's theorem ,we know that

$P(E_1/A)=\large\frac{P(E_1)P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}$

$\Rightarrow \large\frac{1/3\times 1}{1/3\times 1+1/3\times 0+1/3\times 1/2}$

$\Rightarrow \large\frac{2}{3}$

Hence the required probability is $\large\frac{2}{3}$