This is indeed a really good question!!!

Here the eqn you should use is kt=2.303log((a+x)/a)........(1)

[This is a different eqn applied in this case only]

and

log(k₂/k₁)=E_a(T₂-T₁)/(2.303T₁T₂)........(2)

Also see that the milk sours implies that the no. of bacterias should be same.

ie log((a+x)/a) should be same.

At T₁=293K ,

k₂*64=2.303log((a+x)/a)......(3)

At T₂= 276K

k₁*64*3=2.303log((a+x)/a)

So we get k₂/k₁=3

Use (2)

So log(3)=E_a(293-276)/(2.303*293*276)........(4)

At T₃=313K

log(k₃/k₂)=E_a(313-293)/(2.303*293*313).....(5)

(4)/(5) gives

k₃=3.125k₂

Now

k₃*t=2.303log((a+x)/a)

or

3.125k₂*t=2.303log((a+x)/a)

Divide this by (3)

We get t=20.48hrs

I hope you will be clear!!!