Step 1: Idea of apparent depth.
Looking straight down through a transparent slab, an object seems raised. Each slab gives an apparent depth $d_{\text{app}} = \dfrac{d_{\text{real}}}{\mu}$, and with two stacked media the apparent depths simply add.
Step 2: Layers present.
The coin sits below $10\,\text{cm}$ of glass ($\mu = 1.5$) which is itself below $10\,\text{cm}$ of water ($\mu = 4/3$).
Step 3: Apparent depth through the glass.
\[ d_1 = \frac{10}{1.5} = \frac{10}{3/2} = \frac{20}{3} \approx 6.67\,\text{cm}. \]
Step 4: Apparent depth through the water.
\[ d_2 = \frac{10}{4/3} = \frac{30}{4} = 7.5\,\text{cm}. \]
Step 5: Add the two contributions.
\[ d_{\text{total}} = d_1 + d_2 = 6.67 + 7.5 = 14.17\,\text{cm}. \]
Step 6: Round to the option.
The apparent depth of the coin is about $14.2\,\text{cm}$.
\[ \boxed{14.2\ \text{cm}} \]