In groundwater hydrology, the hydraulic head $h$ at a distance $r$ from a pumping well is calculated by the following equation:

$h=\frac{Q}{2T}ln(\frac{2r}{d})+h_w$

where $Q$ is the pumping rate, $T$ is the Transmissivity of the aquifer, $d$ is the diameter of the well, and $h_w$ is the water level in the pumping well.

Given the following information, find $T,$ the transmissivity in $m^2/$ day:

Pumping rate is $2\mathrm{.}1m^3$ per day;

Well diameter is $15cm.$

Water level in the pumping well is $8\mathrm{.}90m$

Hydraulic head measured $97m$ away is $9\mathrm{.}86m$