After how many half-lives a radioactive substance will decay to 6.25% of its original value:

The correct option is Four half lives.

Let's understand how a radioactive substance decays over time step-by-step. A half-life is the time required for a quantity of a radioactive substance to reduce to half of its initial value.

Let the initial amount of the radioactive substance be $N_0$, which represents 100% of its original value.

After one half-life, the remaining amount of the substance is:
$N_1=\frac{1}{2}{N}_0=50\%$ of the original value.

After two half-lives, the remaining amount is halved again:
$N_2=\frac{1}{2}{N}_1=\frac{1}{4}$ of the original value.

After three half-lives, the remaining amount is halved once more:
$N_3=\frac{1}{2}{N}_2=\frac{1}{8}$ of the original value.

After four half-lives, the remaining amount becomes:
$N_4=\frac{1}{2}{N}_3=\frac{1}{16}$ of the original value.

In general, the fraction of the substance remaining after $n$ half-lives is given by the formula:
$N=N_0\cdot {\left(\frac{1}{2}\right)}^n$

For the substance to decay to 6.25% (or a fraction of 0.0625) of its original value:
${\left(\frac{1}{2}\right)}^n=0.0625$

Since $0.0625=\frac{1}{16}={\left(\frac{1}{2}\right)}^4$, we find:
$n=4$

Thus, it takes exactly four half-lives for the radioactive substance to decay to 6.25% of its original value.