Answer
$T \approx 2.1*10^{-2}$ sec
Work Step by Step
Given:
The period is changing at the rate of
$1.26*10^{-5} \frac{sec}{year}$
So, the original period, assuming the period of 0.033 was measured in 2019, is:
T = Original Period - rate*acceleration = $0.033 - $($1.26$ * $10^{-5}$*($2019-1054$)) = $T \approx 2.1*10^{-2}$ sec
We use $\approx$, as we are only given three significant figures.
