Answer
$X_0^2\gt X_α^2$: null hypothesis is rejected.
There is enough evidence to conclude that the standard deviation of the amount of the active ingredient is greater than 5 mg.
Work Step by Step
$H_0:~σ=5$ versus $H_1:~σ\gt5$
$X_0^2=\frac{(n-1)s^2}{σ_0^2}=\frac{(30-1)7.3^2}{5^2}=61.8164$
Right-tailed test:
$n=30$
$d.f.=n-1=29$
$X_α^2=X_{0.1}^2=39.087$
(According to Table VII, for d.f. = 29 and area to the right of critical value = 0.10)
Since $X_0^2\gt X_α^2$, we reject the null hypothesis.
