(C) The isotope $_{88}Ra^{226}$ is radioactive.For this nucleus,the number of protons $(p)$ is $88$ and the number of neutrons $(n)$ is $226 - 88 = 13

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(C) The isotope $_{88}Ra^{226}$ is radioactive.For this nucleus,the number of protons $(p)$ is $88$ and the number of neutrons $(n)$ is $226 - 88 = 138$.The $\frac{n}{p}$ ratio is $\frac{138}{88} \approx 1.568$.Since the $\frac{n}{p}$ ratio is significantly greater than $1.5$,the nucleus is unstable and undergoes radioactive decay.