A first order reaction was followed. The values of ln([R]/(mol/L)) at 0, 10, 20 and 30 min are…

Step 1. The slope. The value falls by $0.30$ in every $10\ \text{min}$. So the slope is $-0.030\ \text{min}^{-1}$ and $k = 0.030\ \text{min}^{-1}$.

Step 2. The start. $\ln[R]_0 = -1.20$, so $[R]_0 = e^{-1.20}\ \text{mol/L} = 0.301\ \text{mol/L}$.

Step 3. At $45\ \text{min}$. $\ln[R] = -1.20 - 0.030 \times 45 = -2.55$. Then $[R] = e^{-2.55}\ \text{mol/L} = 0.0781\ \text{mol/L}$.

Step 4. The half-life. $t_{1/2} = \dfrac{0.693}{0.030\ \text{min}^{-1}} = 23.1\ \text{min}$.