(N/A) The decay follows first-order kinetics,where the rate constant $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{28.1} \ y^{-1}$.
Using the first-order integrated rate equation: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$.
For $t = 10 \ years$:
$10 = \frac{2.303}{0.693 / 28.1} \log \frac{1}{[R]}$
$\log [R] = -\frac{10 \times 0.693}{2.303 \times 28.1} \approx -0.1071$
$[R] = \text{antilog}(-0.1071) \approx 0.7814 \ \mu g$.
For $t = 60 \ years$:
$60 = \frac{2.303}{0.693 / 28.1} \log \frac{1}{[R]}$
$\log [R] = -\frac{60 \times 0.693}{2.303 \times 28.1} \approx -0.6425$
$[R] = \text{antilog}(-0.6425) \approx 0.2278 \ \mu g$.
Thus,$0.7814 \ \mu g$ remains after $10 \ years$ and $0.2278 \ \mu g$ remains after $60 \ years$.