Answer: C

Small Hint:

Write $n=2^a3^bq$, where $q$ is relatively prime to $6$, and let $t$ be the number of divisors of $q$

Big Hint:

Use the two divisor-count equations to restrict $t$ to a divisor of both $28$ and $30$

Solution:

Write $n=2^a3^bq$ with $\gcd(q,6)=1,$ and let $t=\tau(q).$ Then $$\begin{aligned} (a+2)(b+1)t&=28,\\ (a+1)(b+2)t&=30. \end{aligned}$$ Hence $t$ is $1$ or $2.$ Checking the factor pairs of $14,15$ gives no solution when $t=2.$ Checking those of $28,30$ when $t=1$ gives the unique solution $a=5,b=3.$ Therefore $$\begin{aligned} \tau(6n)&=(a+2)(b+2)t\\ &=7\cdot5=35, \end{aligned}$$ so the correct answer is C.