A small telescope has an aperture of diameter 0.15 m…

Step 1. The angle. $\theta_{\min} = \dfrac{1.22 \times 600 \times 10^{-9}\ \text{m}}{0.15\ \text{m}} = 4.88 \times 10^{-6}\ \text{rad} = 4.88\ \mu\text{rad}$.

Step 2. In arcseconds. $\dfrac{4.88\ \mu\text{rad}}{4.848\ \mu\text{rad}} = 1.01$ arcseconds.

Step 3. The new aperture. $0.10$ arcsecond is $0.10 \times 4.848 \times 10^{-6}\ \text{rad} = 0.485 \times 10^{-6}\ \text{rad}$. We want $D$, so $D = \dfrac{1.22\lambda}{\theta_{\min}} = \dfrac{1.22 \times 600 \times 10^{-9}\ \text{m}}{4.848 \times 10^{-7}\ \text{rad}} = 1.51\ \text{m}$.