What is the minimum possible number of satellites serving B exclusively?

From the ratio of satellites serving B, C and S, $B_{total}:C_{total}:S_{total}=2:1:1.$

Hence, $C_{total}=S_{total}.$

Now, $C_{total}=c+x+z+100$

and $S_{total}=s+y+z+100.$

Also, $c=s$ (given in the question).

Therefore, $c+x+z+100=s+y+z+100$

which simplifies to ${x=y.}$

Using $B_{total}=2C_{total},$

we get $b+x+y+100=2(c+x+z+100).$

Since $x=y,$ the $x$-terms cancel, giving ${b=2c+2z+100.}$

Now use the given relation $c=30\%\text{ of }b$, i.e., $c=0.3b.$

Substituting $b=2c+2z+100$, $c=0.3(2c+2z+100).$

Solving, ${c=1.5z+75.}$

Substituting back, ${b=5z+250.}$

The total number of satellites is 1600.

Hence, $1600=b+c+s+x+y+z+100+O.$

Using $s=c,\qquad y=x,\qquad O=z,$

we obtain $1600=b+2c+2x+2z+100.$

Substituting the expressions for $b$ and $c$,

$1600=(5z+250)+2(1.5z+75)+2x+2z+100.$

Simplifying, $1600=10z+500+2x.$

Therefore, ${x=550-5z.}$

From the solution, Since every region represents a number of satellites, $x\ge0.$

Using $x=550-5z,$ we get $550-5z\ge0$ or ${0\le z\le110.}$

Also, $b=5z+250.$

Since $b$ increases as $z$ increases, the smallest possible value of $b$ occurs when $z=0.$

Thus, ${b_{\min}=250.}$