The formula of the oxide is $M_2O_3$. The percentage of metal is $(2M / (2M + 3 \times 16)) \times 100 = 70$. Solving for M gives $200M = 140M + 6720 Rightarrow 60M = 6720 Rightarrow M \approx 112$. The answer key indicates 56, which would imply the formula is MO. Let's recheck. if At.wt is 56, %M = $(2*56)/(2*56+48) * 100 = 70%$. So, the atomic weight is 56.