Comparing with the recurrence form of Master Theorem $T(n) = aT(n/b) + f(n)$
$a=1, b=3, \log_b a=0$
So $n^{\log_b a} = n^0$
$f(n)=cn$
So, $f(n)=\Omega(n^{\log_b a+\epsilon})$ holds for any $\epsilon < 1$ and this is the third case of Master theorem. But Master theorem also requires (only case 3) that regularity condition be satisfied (this ensures that $f(n)$ is still dominant when the recurrence go deeper) which is $af(n/b) \leq df(n)$ for some $d < 1$ and all sufficiently large $n.$ (Using $d$ here as $c$ is already used in $f(n))$
Here we get, $f(n/3) \leq df(n)$
$\implies cn/3 \leq dcn \implies 1/3 \leq d$
Thus we can use any $1/3 \leq d < 1$ to satisfy the regularity condition and Master theorem case 3 is satisfied. Now, applying case $3$ we get
$T(n) = \Theta(f(n)) = \Theta(n)$
answer is A.