The internal energy of a monoatomic gas is 3 nRT. One mole of...

The question states the internal energy of the gas is $U = 3nRT$.
The change in internal energy is $\Delta U = 3nR\Delta T$.
Given: $n = 1$ mole, $\Delta T = 4^{\circ}C = 4$ K, and $R \approx 8.314$ J/mol·K.
$\Delta U = 3 \times 1 \times 8.314 \times 4 = 99.768$ J.
According to the first law of thermodynamics, $\Delta Q = \Delta U + W$.
Given heat supplied $\Delta Q = 126$ J.
Work done by the gas $W = \Delta Q - \Delta U = 126 - 99.768 = 26.232$ J.
Since the piston is light and movable, the process occurs at constant atmospheric pressure $P = 10^5$ Pa.
Work done $W = P \Delta V = P(A \times d)$, where $A$ is the cross-section area and $d$ is the displacement of the piston.
$A = 17 \text{ cm}^2 = 17 \times 10^{-4} \text{ m}^2$.
$26.232 = 10^5 \times (17 \times 10^{-4}) \times d$.
$26.232 = 170 \times d$.
$d = \frac{26.232}{170} \approx 0.1543$ m.
Converting to cm: $d \approx 15.43$ cm.
Rounding to the nearest option, we get $15.5$ cm.