A lead plate 3.0 cm thick has a face of 0.15 m²…

Step 1. Stress. $\dfrac{3.0 \times 10^{4}\ \text{N}}{0.15\ \text{m}^2} = 2.0 \times 10^{5}\ \text{Pa}$.

Step 2. Angle. $\theta = \dfrac{2.0 \times 10^{5}\ \text{Pa}}{5.6 \times 10^{9}\ \text{Pa}} = 3.57 \times 10^{-5}\ \text{rad}$.

Step 3. Shift. $\theta h = 3.57 \times 10^{-5} \times 0.030\ \text{m} = 1.07 \times 10^{-6}\ \text{m}$.