Your conjecture is true. Notice first that it is enough to show that $C=p^k$ for some prime $p$. Indeed, in this case$$B=\sigma(C)=\frac{p^{k+1}-1}{p-1},$$hence$$\mathrm{rad}(C)(B-C)=p\frac{p^k-1}{p-1}=B-1.$$This means that $\varphi(B)=B-1$, so $B$ is prime.
To show that $C$ must be a prime power, assume the contrary. Let $p$ be the least prime factor of $C$. Since $C$ has at least one larger prime factor (otherwise $C=p^k$) , we have$$\mathrm{rad}(C)\geq p(p+1)=p^2+p.$$On the other hand,$$B\geq \varphi(B)=\mathrm{rad}(C)(B-C)\geq (p^2+p)(B-C).$$Therefore,$$C\geq B\left(1-\frac{1}{p^2+p}\right).$$Next, if $C=p_1^{k_1}\ldots p_l^{k_l}$, then$$\sigma(C)=\prod_i \frac{p_i^{k_i+1}-1}{p_i-1}$$and$$\frac{\sigma(C)}{C}=\prod_i\frac{p_i-p_i^{-k_i}}{p_i-1}.$$Every factor in the product above is $\geq 1$, so we can bound the product from below by any one factor. In particular, for some $k\geq 1$$$\frac{\sigma(C)}{C}\geq \frac{p-p^{-k}}{p-1}.$$So$$B=\sigma(C)\geq C\frac{p-1/p}{p-1}=C\frac{p+1}{p}.$$Substituting into the previous inequality, we obtain$$C\geq B\left(1-\frac{1}{p^2+p}\right)\geq C\left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}.$$Dividing by $C$, we arrive at$$1\geq \left(1-\frac{1}{p^2+p}\right)\frac{p+1}{p}=1+\frac{1}{p}-\frac{1}{p^2}>1,$$which is a contradiction.