A reaction of one reactant is run twice at the same temperature…

Step 1. The order. The start concentration doubled and the half-life doubled. A half-life that is proportional to $[R]_0$ belongs to a zero order reaction.

Step 2. The rate constant. $k = \dfrac{[R]_0}{2t_{1/2}} = \dfrac{0.24\ \text{mol/L}}{2 \times 30\ \text{min}} = 4.0 \times 10^{-3}\ \text{mol L}^{-1}\,\text{min}^{-1}$.

Step 3. Time to finish. $t = \dfrac{0.48\ \text{mol/L}}{4.0 \times 10^{-3}\ \text{mol L}^{-1}\,\text{min}^{-1}} = 120\ \text{min}$. This is $2 \times 60\ \text{min}$, twice the half-life of that run.