The period of a pendulum is $T = 2\pi\sqrt{L/g}$. The acceleration due to gravity $g$ decreases with altitude $h$ according to $g_h \approx g_0(1 - 2h/R_E)$, where $R_E$ is the Earth's radius. The ratio of the new period $T_h$ at height $h$ to the original period $T_0$ is:

$$ \frac{T_h}{T_0} = \sqrt{\frac{g_0}{g_h}} \approx \sqrt{\frac{1}{1 - 2h/R_E}} \approx \left(1 - \frac{2h}{R_E}\right)^{-1/2} $$

Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x= -2h/R_E$:

$$ \frac{T_h}{T_0} \approx 1 + \left(-\frac{1}{2}\right)\left(-\frac{2h}{R_E}\right) = 1 + \frac{h}{R_E} $$

The fractional change in period is $(T_h-T_0)/T_0 = h/R_E$. The total time lost, $\Delta t_{lost}$, in a time interval $\Delta t$ is this fraction multiplied by the interval:

$$ \Delta t_{lost} \approx \Delta t \frac{h}{R_E} $$

Substituting the values $\Delta t = 24 \text{ hours} = 86400$ s, $h = 200$ m, and $R_E \approx 6.4 \times 10^6$ m:

$$ \Delta t_{lost} \approx 86400 \text{ s} \times \frac{200 \text{ m}}{6.4 \times 10^6 \text{ m}} $$