1. Given data

The six observations are $1,2,4,5,x,y$ with mean $=5$ and variance $=10$.

2. Use the mean

For 6 observations with mean 5, $\frac{1+2+4+5+x+y}{6}=5$

So, $1+2+4+5+x+y=30$ $12+x+y=30$ $x+y=18$

3. Use the variance

Variance is $\frac{1}{6}\sum (x_i-\bar x)^2=10$ with mean $\bar x=5$.

So, $\frac{1}{6}\left[(1-5)^2+(2-5)^2+(4-5)^2+(5-5)^2+(x-5)^2+(y-5)^2\right]=10$

Compute known terms: $(-4)^2+(-3)^2+(-1)^2+0^2=16+9+1+0=26$

Hence, $\frac{26+(x-5)^2+(y-5)^2}{6}=10$ $26+(x-5)^2+(y-5)^2=60$ $(x-5)^2+(y-5)^2=34$

Now expand: $x^2-10x+25+y^2-10y+25=34$ $x^2+y^2-10(x+y)+50=34$

Using $x+y=18$, $x^2+y^2-180+50=34$ $x^2+y^2=164$

Now, $(x+y)^2=x^2+y^2+2xy$ $18^2=164+2xy$ $324=164+2xy$ $2xy=160$ $xy=80$

So $x,y$ are roots of $t^2-18t+80=0$ $t^2-18t+80=(t-10)(t-8)=0$

Thus, $x=10,\quad y=8$ (or vice versa).

4. Compute mean deviation about the mean

Mean deviation about mean $=\dfrac{1}{6}\sum |x_i-5|$.

The observations are $1,2,4,5,8,10$.

Their absolute deviations from 5 are: $|1-5|=4,\quad |2-5|=3,\quad |4-5|=1,\quad |5-5|=0,\quad |8-5|=3,\quad |10-5|=5$

Sum: $4+3+1+0+3+5=16$

Therefore, $\text{Mean deviation about mean}=\frac{16}{6}=\frac{8}{3}$

5. Option check

$\boxed{\frac{8}{3}}$ So the correct option is B.