1. Write the six terms of the A.P.

Let the six numbers in A.P. be $a_1=a,\, a_2=a+d,\, a_3=a+2d,\, a_4=a+3d,\, a_5=a+4d,\, a_6=a+5d.$

2. Use the condition $a_1+a_3=10$

We have $a_1+a_3=a+(a+2d)=2a+2d=10.$ So, a+d=5. \tag{1}

3. Use the mean of the six numbers

The mean of six terms of an A.P. is the average of first and last term: $\bar x=\frac{a_1+a_6}{2}=\frac{a+(a+5d)}{2}=a+\frac{5d}{2}.$ Given mean is $\frac{19}{2}$, so a+\frac{5d}{2}=\frac{19}{2}. \tag{2}

Multiply by $2$: 2a+5d=19. \tag{3}

From (1), $a=5-d.$ Substitute into (3): $2(5-d)+5d=19$ $10-2d+5d=19$ $3d=9$ $d=3.$ Then $a=5-3=2.$

So the six numbers are $2,5,8,11,14,17.$

4. Compute the variance

Mean is $\bar x=\frac{19}{2}=9.5.$

Deviations from mean: $2-9.5=-7.5,\quad 5-9.5=-4.5,\quad 8-9.5=-1.5,$ $11-9.5=1.5,\quad 14-9.5=4.5,\quad 17-9.5=7.5.$

Squares: $(-7.5)^2=56.25,\quad (-4.5)^2=20.25,\quad (-1.5)^2=2.25,$ $(1.5)^2=2.25,\quad (4.5)^2=20.25,\quad (7.5)^2=56.25.$

Sum of squares: $56.25+20.25+2.25+2.25+20.25+56.25=157.5.$

Variance is $\sigma^2=\frac{157.5}{6}=26.25=\frac{105}{4}.$

Therefore, $8\sigma^2=8\cdot \frac{105}{4}=210.$

5. Check options

Thus the correct option is $\boxed{\text{B: }210}.$