Answer
$\approx 69\;\rm g$
Work Step by Step
The ball will reach its terminal speed when its gravitational force (downward) and the drag force (upward) are equal and then the net force exerted on it is zero.
$$\sum F_y=D-mg=0$$
Thus,
$$D=mg$$
and we know that the drag force is given by
$$D=\frac{1}{2}C\rho A v^2$$
where $C$ is the e drag coefficient, $\rho$ is the air density, $A$ is the cross-sectional area of the object that faces the air, $v$ is the falling object speed; and when the net force is zero, $v=v_T$.
Hence,
$$ \frac{1}{2}C\rho A v_T^2=mg$$
Therefore,
$$m=\frac{C\rho A v_T^2}{2g}\tag 1$$
Now we need to find the cross-sectional area of the ball which is the area of its central circle, see the figure below.
$$A=\pi r^2=\pi \left[\dfrac{L}{2}\right]^2$$
where $L$ is the diameter.
Plugging into (1);
$$m=\frac{C\rho v_T^2}{2g} $$
$$m=\frac{C\rho \left(\pi \left[\dfrac{L}{2}\right]^2 \right) v_T^2}{2g}$$
$$m=\frac{\pi C\rho L^2 v_T^2}{8g}$$
Plugging the known;
$$m=\frac{\pi \cdot 0.5\cdot 1.2\cdot (6.5\times10^{-2})^2\cdot 26^2}{8\cdot 9.8}=6.867\times10^{-2}\;\rm kg$$
$$m=\color{red}{\bf68.67}\;\rm g$$
