Since in $t=2$s, distance covered $s=2m$ we have ,
$s=ut+\large\frac{1}{2}$$at^2$
$2= u \times 2+\large\frac{1}{2}$$a \times 2^2$
or $u+a=1$-----(i)
As the distance covered in $t=2$s to $6\;s$ is $2.2 m$ We have
$2.2=u(6-2)+\large\frac{1}{2}$$a (6^2-2^2)$
$2u+8a=1.1$------(ii)
From equation (i) and (ii)
$u=1.15ms^{-1}$
$a= -0.15 ms^{-2}$
From the relation $v=u+at$, We have on putting,
$u=1.15 ms^{-1},a = -0.15 ms^{-2}$ and $ t=7\;s$
$v= 1.15 +(-0.15) \times 7$
$\quad=0.10 ms^{-1}$
Hence B is the correct answer.