The standard value of acceleration due to gravity is taken as 10...

Using the principle of dimensional homogeneity, the magnitude of a physical quantity remains the same in different systems of units: $n_1 u_1 = n_2 u_2$.

For acceleration, the dimensional formula is $[LT^{-2}]$.

Given:
$n_1 = 10$, $u_1 = \text{m/s}^{2}$
$n_2 = 129600$
$L_2 = 1 \text{ km} = 1000 \text{ m}$

Equating the two:
$n_1 [L_1 T_1^{-2}] = n_2 [L_2 T_2^{-2}]$
$10 \times \left(\dfrac{1 \text{ m}}{1 \text{ s}^{2}}\right) = 129600 \times \left(\dfrac{1000 \text{ m}}{T_2^{2}}\right)$

Rearranging to solve for $T_2^{2}$:
$T_2^{2} = \dfrac{129600 \times 1000}{10} \text{ s}^{2}$
$T_2^{2} = 12960000 \text{ s}^{2}$

Taking the square root:
$T_2 = 3600 \text{ s}$

Since $1 \text{ hour} = 3600 \text{ s}$, the new unit of time is $1 \text{ hour}$.

Answer: $1 \text{ hour}$