Total students = 100

At t = 0 (zero day), infected student = 2

Let at t = t day infected student = x

$$\therefore$$ At t = t day non infected student = (100 $$-$$ x)

Rate of infection $$ = {{dx} \over {dt}}$$

Given, $${{dx} \over {dt}} \propto x(100 - x)$$

$$ \Rightarrow \int\limits_{}^{} {{{dx} \over {x(100 - x)}} = \int\limits_{}^{} {k\,dt} } $$

$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {{{100 - x + x} \over {x(100 - x)}}dx = k\,t + c} $$

$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {\left( {{1 \over x} + {1 \over {100 - x}}} \right)dx = k\,t + c} $$

$$ \Rightarrow {1 \over {100}}\left[ {\ln x - \ln (100 - x)} \right] = k\,t + c$$

$$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} = k\,t + c$$ ...... (1)

Given, At, t = 0, x = 2

$$\therefore$$ $${1 \over {100}}\ln {2 \over {98}} = c$$

Putting value of c in equation (1), we get

$${1 \over {100}}\ln {x \over {100 - x}} = kt + {1 \over {100}}\ln {2 \over {98}}$$

$$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} - {1 \over {100}}\ln {2 \over {98}} = kt$$

$$ \Rightarrow {1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = kt$$

Given, At t = 4, x = 30

$$\therefore$$ $${1 \over {100}}\ln {{30 \times 98} \over {2(70)}} = k \times 4$$

$$ \Rightarrow k = {1 \over {400}}\ln 21$$

$$\therefore$$ $${1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = t \times {1 \over {400}} \times \ln 21$$

Now, when t = 8, then r = ?

$${1 \over {100}}\ln {{49x} \over {(100 - x)}} = 8 \times {1 \over {400}} \times \ln 21$$

$$ \Rightarrow \ln {{49x} \over {(100 - x)}} = 2\ln 21$$

$$ \Rightarrow {{49x} \over {100 - x}} = {21^2}$$

$$ \Rightarrow {x \over {100 - x}} = {{21 \times 21} \over {49}}$$

$$ \Rightarrow {x \over {100 - x}} = 9$$

$$ \Rightarrow x = 900 - 9x$$

$$ \Rightarrow 10x = 900$$

$$ \Rightarrow x = 90$$