Here is what I did to start this problem:
Let A be the first attack. Let $X_1, X_2, ... $ be the i.i.d. arrival sequence of attacks. Then, $A = \sum_{i=1}^{k} X_i$ for each $k=1, ..., 5$. Let $Z_k$ be the time of the second attack. Then, $A = min_{1\leq k \leq 5}Z_k$ and $Z_k $ ~ $\Gamma (2, \frac{1}{5})$ with density $f(t) = \frac{te^{\frac{-t}{5}}}{5^2}$.
I don't know if this is the right way to approach this problem because it doesn't consider the random density function for the amount of time that the PC is powered on. I would appreciate any help either correcting my reasoning or showing me another way to approach the problem.
The number of virus attacks are dependent on the time that the PC is on, thus the distribution of the virus attacks is
$$V|T\sim Po\left(\frac{t}{5}\right)$$
and thus
$$f_{VT}(v,t)=\frac{e^{-t/5}\left(\frac{t}{5}\right)^v}{v!}\cdot\frac{5}{4t^2}$$
The probability that $V=2$ is
$$\frac{1}{40}\int_1^5 e^{-t/5}dt=\frac{5e^{-1/5}-5e^{-1}}{40}\approx 5.64\%$$
The correctness of the solution can be verified with a calculator that gives
$\mathbb{P}[V=0]\approx68.06\%$
$\mathbb{P}[V=1]\approx 25.08\%$
$\mathbb{P}[V=2]\approx 5.64\%$
$\mathbb{P}[V=3]\approx 1.03\%$
$\mathbb{P}[V=4]\approx 0.16\%$
that is approx p=1.