How do i convert a boolean expression to NOR expression?
There is a trick for converting sum-of-products to nand gates and a trick for converting product-of-sums to nor gates.  (In diagram below).

But you want nor gates and you are starting in sum-of-products form.  So how do you convert sum-of-products to product-of-sums?  (Also this one.)

Well, in your case the product-of-sums form is trivially derived just by factoring out the \$\overline{B}\$ to get \$\overline{B}(\overline{A}+C)\$.  More generally though, the plug-n-chug way you convert sum-of-products to product-of-sums is


double negate your formula.  \$\overline{\overline{\overline{A} \, \overline{B} + \overline{B} \, C}} \$
Use Demorgan to push the first not through the sum. \$\overline{(\overline{\overline{A}\,\overline{B}})\,(\overline{\overline{B}\,C})} \$
And one more Demorgan to turn the products into sums. \$ \overline{(A+B)(B+\overline{C})}  \$
Now multiply the product-of-sums to get a negated sum-of-products. \$ \overline{AB+A\overline{C}+B+B\overline{C}} = \overline{A\overline{C} + B}\$
Almost there!  A negated sum of products is a product of sums by two more Demorgans: \$ (\overline{A\,\overline{C}})(\overline{B}) = (\overline{A}+C)(\overline{B}).\$


Now you've got a product of sums.

Finally, here's how you convert product-of-sums to nor gates.  It's sometimes called "pushing bubbles".







simulate this circuit &ndash; Schematic created using CircuitLab

So I get what you get (and what @jippie got).  (\$\overline{A}\$ NOR C) NOR B.  \$\overline{\overline{\overline{A+B}+C}+B}\$ is logicially equivalent, of course, but requires 3 2-input nor gates instead of 2 2-input nor gates and a "1-input nor gate" (not gate).