1. Use the radioactive decay relation

For activity of a radioactive substance:

$A = A_0 e^{-\lambda t}$

Also, half-life is related to decay constant by:

$t_{1/2} = \frac{0.693}{\lambda}$

Given:

$t_{1/2} = 200\ \text{days}$

So,

$\lambda = \frac{0.693}{200}$

2. Find the fraction of activity remaining after 83 days

$\frac{A}{A_0} = e^{-\lambda t} = e^{-\left(\frac{0.693}{200}\right)(83)}$

$\frac{A}{A_0} = e^{-0.287595}$

Now convert to base 10 using:

$e^{-x} = 10^{-x/2.303}$

So,

$\frac{A}{A_0} = 10^{-0.287595/2.303}$

$\frac{A}{A_0} = 10^{-0.125}$

Given:

$\text{antilog } 0.125 = 1.333$

Hence,

$10^{-0.125} = \frac{1}{10^{0.125}} = \frac{1}{1.333} \approx 0.75$

3. Convert to percentage

$\%\text{ activity remaining} = 0.75 \times 100 = 75\%$

4. Nearest integer

$\boxed{75}$

5. Comparison with stored answer

Stored correct answer = $75$

Our derived answer also = $75$

So the answer matches.