You could do it "manually" by converting it yourself
byte b = 0, pot = 1;
for (int i = 5; i >= 0; i--) {
// -48: the character '0' is No. 48 in ASCII table,
// so substracting 48 from it will result in the int value 0!
b += (str.charAt(i)-48) * pot;
pot <<= 1; // equals pot *= 2 (to create the multiples of 2 (1,2,3,8,16,32)
}
This will multiply the bits with (1, 2, 4, 8, 16, 32) in order to determine the resulting decimal number.
Another possibility would be to really manually calculate the 6 binary digits into a decimal value:
byte b = (byte)
((str.charAt(5) - '0') * 1 +
(str.charAt(4) - '0') * 2 +
(str.charAt(3) - '0') * 4 +
(str.charAt(2) - '0') * 8 +
(str.charAt(1) - '0') * 16 +
(str.charAt(0) - '0') * 32);